Revised list of TBCs

1 files changed, 0 insertions, 185 deletions

diff --git a/Aircraft_Propulsion_by__S._Farokhi/Chapter6.ipynb b/Aircraft_Propulsion_by__S._Farokhi/Chapter6.ipynb
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-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:9667e81e13ddfe520a56dd149f9bdf0ca5ae444b71480274d3f1dd69a833ceb1"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter6-Combustion Chambers and After burners"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex1-pg309"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#determine number of mole of hydrogen and oxygen\n",

-      "nH2=12/2. ##molecular mass og hydrogen =2kg/kmol\n",

-      "nO2=8/32. ##Molecular mass of O2=32kg/kmol\n",

-      "print'%s %.f %s'%(\"No. of kilomoles of H2\",nH2,\"\")\n",

-      "print'%s %.2f %s'%(\"No. of kilomoles of O2\",nO2,\"\")"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "No. of kilomoles of H2 6 \n",

-        "No. of kilomoles of O2 0.25 \n"

-       ]

-      }

-     ],

-     "prompt_number": 2

-    },

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Ex3-pg317"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "#calculate lower and higher heating values of hydrogen\n",

-      "T=298.16 ##in K\n",

-      "dhf=-241827. ##heat of formation of H2O(g in kJ.\n",

-      "n=1 ##kmol\n",

-      "Qr=n*dhf ##kJ/kmol\n",

-      "LHV=(-1.)*Qr/2.\n",

-      "print'%s %.1f %s'%(\"LHV in\",LHV,\"kJ/kg\")\n",

-      "HHV=LHV+9*2443\n",

-      "print'%s %.1f %s'%(\"HHV in \",HHV,\"kJ/kg\")"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "LHV in 120913.5 kJ/kg\n",

-        "HHV in  142900.5 kJ/kg\n"

-       ]

-      }

-     ],

-     "prompt_number": 3

-    },

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Ex5-pg320"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "#calcualte the ratio Nh2/no2 of the reactants and fuel oxdizer and adiabatic flame temperature\n",

-      "##from equation CH4+2.4(O2+3.76N2)-->CO2+2H2O+0.4O2+9.02N2\n",

-      "f=(12+4.)/(2.4*(32.+3.76*28.)) ##fuel to air ratio based on mass.\n",

-      "fs=(12+4.)/(2.*(32.+3.76*28.)) ##fuel to air ratio based on stoichometric condition.\n",

-      "feq=f/fs\n",

-      "print'%s %.7f %s'%(\"fuel to air ratio based on mass\",f,\"\")\n",

-      "print'%s %.7f %s'%(\"fuel to air ratio based on stoichometric condition\",fs,\"\")\n",

-      "print'%s %.7f %s'%(\"Equivalent ratio\",feq,\"\")\n",

-      "dH=-802303 ##kJ\n",

-      "dC=484.7 ##kJ\n",

-      "Dt=(-1)*dH/dC ##Dt=T2-Tf\n",

-      "Tf=25+273\n",

-      "T2=Dt+Tf\n",

-      "print'%s %.4f %s'%(\"The diabatic flame temperature in\",T2,\" K\")"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "fuel to air ratio based on mass 0.0485625 \n",

-        "fuel to air ratio based on stoichometric condition 0.0582751 \n",

-        "Equivalent ratio 0.8333333 \n",

-        "The diabatic flame temperature in 1953.2569  K\n"

-       ]

-      }

-     ],

-     "prompt_number": 4

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Ex6-pg323"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "import numpy\n",

-      "#calculate mole fraction of N2 at equlibrium when pm is 1 atm and 10 atms\n",

-      "print(\"Example 6.6\")\n",

-      "Kp=0.1\n",

-      "\n",

-      "pm=1.\n",

-      "y=2\n",

-      "d=numpy.roots(y)\n",

-      "x=0.1561738 \n",

-      "print'%s %.2f %s '%(\"(a)Mole fraction of N2 at equilibrium when pm is 1 atm:\",x,\"\")\n",

-      "#part (b)\n",

-      "Kp=0.1\n",

-      "\n",

-      "pm=10.\n",

-      "x=0.0499376\n",

-      "y=- 0.1 + 40.1*x\n",

-      "d=numpy.roots(y)\n",

-      "i=numpy.linspace(1,2,num=1);\n",

-      "print'%s %.2f %s '%(\"(b)Mole fraction of N2 at equilibrium when pm is 10 atm:\",x,\"\")\n",

-      "    \n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Example 6.6\n",

-        "(a)Mole fraction of N2 at equilibrium when pm is 1 atm: 0.16  \n",

-        "(b)Mole fraction of N2 at equilibrium when pm is 10 atm: 0.05  \n"

-       ]

-      }

-     ],

-     "prompt_number": 26

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
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